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Interference Fringes in a Diffraction envelope

  • #1
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Homework Statement


My question is at https://imgur.com/a/lDOUzjY . The de Broglie wavelength of the atoms in question is 45 pm. The distance from the slits to the detection plane is 1.95m. Each slit width is 1 micrometer. The distance between the slits is 8 micrometers.

2. Homework Equations

y_{min} = λL/a where λ is the wavelength, L is the distance between the slits and the detection plane, and a is the slit width. This is for a minimum in a diffraction experiment.

The Attempt at a Solution


The width of the diffraction envelope is 2y=2(λL/a)=175.5 micrometers where y is the distance from the central maximum to the first minimum(either above or below it). My question is about the number of interference fringes that fit in the envelope. Is it the intersection of the two diffraction envelopes of the two slits and not the single slit envelope of the first part?
 
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Answers and Replies

  • #2
Charles Link
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Usually the letter ## b ## us used for slit width. The diffraction envelope has its zeros when ## m \lambda=b \sin(\theta) ## with integer ## m \neq 0 ##. The question is asking how many maxima do you find between the ## \theta ## corresponding to ## m=-1 ## and ## m=+1 ##. The maxima of the interference pattern are found where ## m \lambda=d \sin(\theta) ##, and this time integer ## m=0 ## is included. Here ## d ## is the distance between slits. ## \\ ## You need to find the largest ## m=m_c ## where ## \theta ## is still less than the place where the diffraction envelope first goes to zero that you found above. Then to count how many maxima you have including ## m=0 ## inside that envelope. That count is a very simple ## 2m_c+1 ##. ## \\ ## And let me give you what may be a helpful hint: I think you will find that ## m=8 ## of the second part lands right on the ## m=1 ## of the first part. Here, that maximum will not appear because the diffraction envelope gets multiplied by the interference pattern intensity result (for very narrow slits or point sources), giving zero intensity there. It should be fairly obvious what ## m_c ## is. And note that the wider the slit, the narrower the diffraction envelope becomes. Point sources and thin slits have very wide diffraction envelopes.
 
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  • #3
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Usually the letter ## b ## us used for slit width. The diffraction envelope has its zeros when ## m \lambda=b \sin(\theta) ## with integer ## m \neq 0 ##. The question is asking how many maxima do you find between the ## \theta ## corresponding to ## m=-1 ## and ## m=+1 ##. The maxima of the interference pattern are found where ## m \lambda=d \sin(\theta) ##, and this time integer ## m=0 ## is included. Here ## d ## is the distance between slits. ## \\ ## You need to find the largest ## m=m_c ## where ## \theta ## is still less than the place where the diffraction envelope first goes to zero that you found above. Then to count how many maxima you have including ## m=0 ## inside that envelope. That count is a very simple ## 2m_c+1 ##. ## \\ ## And let me give you what may be a helpful hint: I think you will find that ## m=8 ## of the second part lands right on the ## m=1 ## of the first part. Here, that maximum will not appear because the diffraction envelope, (the wider the slit, the narrower the diffraction envelope), gets multiplied by the interference pattern intensity result (for very narrow slits or point sources), giving zero intensity there. It should be fairly obvious what ## m_c ## is.
m_c = 7

Thank you
 
  • #4
Charles Link
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m_c = 7
If this is a homework assignment, your teacher will want you to show your work though of what happens for ## m=8 ##.
 
  • #5
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If this is a homework assignment, your teacher will want you to show your work though of what happens for ## m=8 ##.
Why does the diffraction envelope get multiplied by the interference pattern intensity result?
 
  • #6
Charles Link
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Why does the diffraction envelope get multiplied by the interference pattern intensity result?
This follows from a computation/superposition to compute the total electric field, keeping track of the phases at each location including taking account of the phase difference that will occur across the different positions off the individual slits. For the case of slits with identical widths, the computation results in a common factor that occurs in all of the slits which is the diffraction envelope. The remaining factor is the interference pattern factor from the slits being treated as point sources. I will try to find you a "link"... Try this: https://sites.ualberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html
 

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